Packing fraction of NaCl unit cell [assuming ideal] if ions along one of its body diagonal are absent ( ${Cl}^{-}$ forms FCC):

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$P . F . = \frac{\frac{4}{3} π (r_{+}^{3} + r_{+}^{3})}{16 \sqrt{2} r_{+}^{3}}$

$P.F. = \frac{\frac{4}{3} π (4 r_{+}^{3} + \frac{15}{4} r_{-}^{3})}{16 \sqrt{2} r_{-}^{3}}$

$P.F. = \frac{\frac{4}{3} π (3 r_{+}^{3} + 4 r_{-}^{3})}{16 \sqrt{2} r_{-}^{3}}$

$P. F. = \frac{\frac{4}{3} π (3 r_{+}^{3} + \frac{15}{4} r_{-}^{3})}{16 \sqrt{2} r_{-}^{3}}$

answer is B.

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Detailed Solution

An edge diagonal has $1 + \frac{1}{4} + \frac{1}{4} = \frac{3}{2}$ sodium atom misssing.

One unit cell contain $4 - \frac{3}{2} = 2.5$ sodium ion and 4 chloride ions.

$Packing fraction (P . F) = \frac{Volume occupied}{Total volume}$

volume occupied $= \frac{4}{3} π (\frac{5}{2} r_{+}^{3} + 4 r_{-}^{3})$

Totalo voolume $= 16 \sqrt{2} 1_{-}^{3}$

$P. F. = \frac{\frac{4}{3} π (3 r_{+}^{3} + \frac{15}{4} r_{-}^{3})}{16 \sqrt{2} r_{-}^{3}}$

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